Let $X$ be a scheme of finite type over $Spec(\mathbb C)$ and let $i:Y \hookrightarrow X$ be a closed subscheme, defined by a sheaf of ideals $J \subset \mathcal O_X$.

Then there is an induced map $i^ h:Y^h \rightarrow X^h$ between the associated analytic spaces.

First: is this map a closed immersion in the sense of analytic spaces, with which I mean that it gives a homeomorphism on its image and the map on structure sheaves $\mathcal O_{X^{h}} \rightarrow i^h_* \mathcal O_{Y^{h}}$ is an epimorphism?

Second: if the first is true, then: is the ideal sheaf of $Y^h$ identical with $J^{h}$, the analytification of the coherent sheaf $J$ ?

Let me note why I am not able to conclude these things simply from exactness of the analytification functor: One would like to say that $(i_*\mathcal O_Y)^{h} \simeq i^{h}_*\mathcal O_{Y^h}$, but this doesn't seem to be deducable from flat base change because we are only on the level of locally ringed spaces.

So I would appreciate very much any kind of careful help in finding a clean answer to these two questions. Thank you.

Have you had a look at the beginning of Serre's GAGA paper ? There is a very down to earth description of the analytification functor there.
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Damian RösslerApr 11 '13 at 11:00

Yes to the first question. This can be easily seen in affine charts.
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Martin BrandenburgApr 11 '13 at 11:21

I answer the second question myself, as I could find a reference for it: the indicated base change is indeed an iso. The reference is the article "Cohomology of Algebraic Varieties", paragraph 2, Thm. 2.2, "Algebraic Geometry II" (ed.: Shafarevich) where the claim is shown for proper morphisms and coherent sheaves.
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RichardApr 11 '13 at 11:39

1

Thanks to Damian and Martin for their comments!
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RichardApr 11 '13 at 11:39